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First of all, the question is not about the general solution. My question on my book asks me to find the integrating factor.

Suppose I have simple IDE

$$(\sin y)\mathbb dx +(\cos x)\mathbb dy=0$$

I don't know how to find the integrating factor $(\mu)$ manually.

What i've done so far is :

Suppose it has form

$$M(x,y) \mathbb dx+N(x,y) \mathbb dy=0$$

Case 1 : $\mu$ is a function of $x$ only

$$\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\\ &=\frac{\cos y +\sin x}{\cos x} \end{aligned}$$

It doesn't work, cz still remaining the $y$

Case 2 : $\mu$ is a function of $y$ only

$$\begin{aligned} \mu&=\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\\ &=\frac{-\sin x -\cos y}{\sin y} \end{aligned}$$

Still remaining the $x$

Case 3 : $\mu$ is a function of $xy$

$$\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{My - Nx}\\ &=\frac{\cos y +\sin x}{y \sin x-x \sin y} \end{aligned}$$

I have no idea for this one...


But, when i use trial and error, i found

$$\mu=\csc y\cdot \sec x$$

But it's just a guess.

How to find the integrating factor systematically? I mean using such $\mu=e^{\int f(x)\,\mathbb dx}$?

Help me, and thanks.

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  • $\begingroup$ The exponential integral version of the integrating factor, I believe, only works for linear DE's, which this is not. $\endgroup$ – Adrian Keister Sep 27 '19 at 13:27
  • $\begingroup$ Actually i found on youtube that is could be... but i don't know what the reference is. $\endgroup$ – user516076 Sep 27 '19 at 13:30
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    $\begingroup$ There is no general recipe. If you assume $\mu=\mu(x)$, then you get $M_y \mu = N_x \mu + N \mu'$ so that $\ln(\mu)'=\frac{M_y - N_x}{N}$ which only helps if that is just a function of $x$. There's an analogous way to try $\mu$ being a function of just $y$. But if it is a function of both then you have $M_y \mu + M \mu_y = N_x \mu + N \mu_x$ which is too general to be useful without some further idea on what $\mu$ might look like. $\endgroup$ – Ian Sep 27 '19 at 13:36
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    $\begingroup$ That said, a situation like this one where an integrating factor makes it separable is one to keep in mind; that one amounts to requiring $(M\mu)_y=(N\mu)_x=0$. $\endgroup$ – Ian Sep 27 '19 at 13:40
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    $\begingroup$ You can view it as a pure eyeball guess, or you can view it slightly more systematically as "can I make this separable?" and then try to solve $(M\mu)_y=(N\mu)_x=0$. (This particular problem also fits into a general framework of $M(y) dx + N(x) dy = 0 \Rightarrow \mu=(MN)^{-1}$.) $\endgroup$ – Ian Sep 27 '19 at 13:41
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$$\sin y \,dx + \cos x dy =0 ~~~(1)\implies \frac{dx}{\cos x}+ \frac{dy}{\sin y}=0~~~(2).$$ Integrating, we get $$ \ln|\tan(x)+\sec(x)|- \ln|\cot(y)+\csc(y)|=C.$$

Note it is as though $IF = \dfrac{1}{\cos x \sin y}= \sec x \csc y,$ which upon multiplying the ODE (1) makes it EXACT as (2) where $$\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}.$$

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  • $\begingroup$ In this case i don't need the general solution. Bcz it's easy to find this nonlinear DE with separation variable. I just need an integrating factor. Even though "maybe" it gives me the same or "maybe" different solution. Thanks anyway. 1+ $\endgroup$ – user516076 Sep 27 '19 at 13:26
  • $\begingroup$ @user516076 yes, it is simple ODE which is easily separable in $x$ and $y$. $\endgroup$ – Dr Zafar Ahmed DSc Sep 27 '19 at 13:27
  • $\begingroup$ Thanks, but my question on my book asks me the integrating factor... not the solution. :) $\endgroup$ – user516076 Sep 27 '19 at 13:28
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    $\begingroup$ @ user516075interesting you may say that $IF=\frac{1}{\cos x \sin y}$ is the integrating facor which when multiplied to the ODE makes it EXACT as $\frac{\partial M }{\partial y} = \frac{\partial N}{\partial x}.$ So have got the correct IF $\endgroup$ – Dr Zafar Ahmed DSc Sep 27 '19 at 13:33
  • $\begingroup$ YEAH, i just cant explain it well... i know you'll know. :) $\endgroup$ – user516076 Sep 27 '19 at 13:34

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